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JAMB UTME - Physics - 2019

Question 1 Report

A vibrator causes water ripples to travel across the surface of a tank. The wave travels 50cm in 2s and the distance between successive crests is 5cm. Calculate the frequency of the vibrator

5Hz

6Hz

5.2Hz

25Hz

Answer Details

The frequency of the vibrator can be calculated using the formula: frequency = speed / wavelength where speed is the speed of the wave, and wavelength is the distance between successive crests. In this case, we are given that the wave travels 50cm in 2s, which means the speed of the wave is: speed = distance / time = 50cm / 2s = 25cm/s We are also given that the distance between successive crests is 5cm, which is the wavelength. Therefore, the frequency of the vibrator is: frequency = speed / wavelength = 25cm/s / 5cm = 5Hz So the correct answer is 5Hz.

Read lesson note on Motion (WAEC)

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Question 11 Report

The diagram shows a uniform meter rule AB which balances horizontally at the 90cm mark when a mass of 0.2kg is suspended from B. Calculate the mass of the meter rule.

0.05kg

0.20kg

0.80kg

1.00kg

Answer Details

M $_{r}$ (90 - 50) = 0.2(100 - 90)
40M $_{r}$ = 0.2 × 10

M $_{r}$ = $\frac{2}{40}$ = 0.05kg

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 15 Report

The distance between an object and its real image in a convex lens is 40cm. If the magnification of the image is 3, calculate the focal length of the lens

6.5 cm

7.5 cm

8.5 cm

4.5 cm

Answer Details

u + v = 40

$\frac{v}{u}$ = 3

v = 3u

u + 3u = 40

4u = 40

u = 10cm

v = 3u = 30cm

f = $\frac{u v}{u + v} = \frac{10 (30)}{10 + 30} = \frac{300}{40}$

= 7.5 cm

Read lesson note on Gravitational Field (WAEC)

Gravitational Field

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Question 17 Report

A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge

8 x 10

^{- 4}

C

4 x 10

^{- 2}

C

8 x 10

^{- 3}

C

4 x 10

^{- 8}

C

Answer Details

C

_{T}

=

C

_{1}

× C

_{2}

C

_{1}

+ C

_{2}

=	3 × 63 + 6

= $\frac{18}{9}$ = 2μf
Q = CV
⇒ 2 × 10 $^{- 6}$ × 400
⇒ 800 × 10 $^{- 6}$ C = 8 × 10 $^{- 4}$ C

Read lesson note on Capacitors (JAMB)

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Question 19 Report

The height at which the atmosphere cases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is

\((\rho_m = 13.6g/cm^3\ \rho = 0.00013g/cm^3)\)

380mmHg

570mmHg

190mmHg

480mmHg

Answer Details

ρ $_{m}$ h $_{m}$ = ρ $_{a}$ h $_{a}$
13.68(760 - p) × 10 $^{- 3}$ = 13 × 10 $^{- 5}$ (20 × 10 $^{3}$ )

760

-

p

=

13 × 10

^{- 5}

× 20 × 10

^{3}

13.68 × 10

^{- 3}

=

19.00

×

10

^{1}

760 - p = 190
p = 760 - 190 = 570mmHg

Read lesson note on Pressure (JAMB)

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Question 27 Report

A thermocouple thermometer is connected to a millivoltmeter which can read up to 10mV. When one junction is in ice at 0°C and the other is steam at 100°C, the millivoltmeter reads 4mV. What is the maximum temperature which this arrangement can measure

100

^{o}

C

248

^{o}

C

250

^{o}

C

350

^{o}

C

Answer Details

The maximum temperature which this arrangement can measure is 250°C. A thermocouple thermometer works by using the thermoelectric effect, which is the phenomenon that occurs when two dissimilar metals are joined together to form a loop and a temperature difference is established between the two junctions. This temperature difference generates a small electrical voltage, which can be measured using a millivoltmeter. The voltage generated is proportional to the temperature difference between the two junctions. In the case of the thermocouple thermometer described, one junction is in ice at 0°C and the other is steam at 100°C, and the millivoltmeter reads 4mV. This means that the voltage generated by the thermocouple is 4 millivolts, which corresponds to a temperature difference of 100°C. However, the millivoltmeter can only read up to 10mV, so the maximum temperature difference it can measure is 10mV / 4mV/°C = 250°C. This means that the maximum temperature which this arrangement can measure is 250°C.

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 36 Report

A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the body is 5 rad/s. Calculate its speed when it is 6m from the center of the motion.

20 m/s

30 m/s

40 m/s

50 m/s

Answer Details

From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω

\sqrt{A^{2} - X^{2}}

= 5

\sqrt{10^{2} - 6^{2}}

= 40m/s

Read lesson note on Simple Harmonic Motion (WAEC)

Simple Harmonic Motion

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Question 39 Report

A metal rod has a length of 100cm at \(200^\circ C\). At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is \(2 \times 10^{-5}C^{-1}\)

200

^{o}

C

300

^{o}

C

100

^{o}

C

-100

^{o}

C

Answer Details

The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.

Read lesson note on Thermal Expansion (JAMB)

Read lesson note on Temperature And Its Measurement (JAMB)

Thermal Expansion

Temperature And Its Measurement

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